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Proof step 3 from proof inrec (previous step)

∀ a, d, b, d0. a × d ∈ ax0 ∨ a = b → (∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ ax0) ⊦ a × d ∈ p + ax0 ∨ a = b → (∃ d1. a × d1 = b × d0 ∨ a × d1 ∈ p + ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0, d2, b0, d3. a0 × d2 ∈ ax0 ∨ a0 = b0 → (∃ d1. a0 × d1 ∈ ax0 ∨ a0 = b0 ∧ d1 = d3), (a × d ∈ ax0 ∨ a × d = p) ∨ a = b ⊦ ∃ d1. a × d1 = b × d0 ∨ a × d1 = p ∨ a × d1 ∈ ax0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ p ∈ p0 ' + ax ↔ p ≠ p0 ∧ ¬ p ∈ ax;% ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);% ⊦ p + ax = p ' + ax;